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I would like to factor the polynomial $p(x,y)=x^p-y^p$ for some small prime $p$ $(p=3,5,\text{or } 7)$ and for all values of $p(x,y)$ with $1 < x < 1000$ and $1 < y < x$.

There is a similar post, but the answers just indicate that you can factor out $(x-y)$. I am aware of this, but I still cannot find a fast way to factor to rest of the polynomial efficiently. In a previous post I learnt quick ways to factor the polynomial $(x-y)$ using a sieve and I am hoping that there is a similar and efficient way to factor this polynomial.

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  • $\begingroup$ over which field? oh, and please see math.meta.stackexchange.com/questions/5020 $\endgroup$ – Lord Shark the Unknown Jul 8 at 18:04
  • $\begingroup$ @LordSharktheUnknown just over the integers. And thank you! Sometimes I forget to type in Latex haha $\endgroup$ – sqrt-3299 Jul 8 at 18:05
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    $\begingroup$ I want to point out that you're not asking to factor a polynomial, you're asking to factor the outputs of the polynomial for a range of values. That is a different question. $\endgroup$ – runway44 Jul 11 at 20:45
  • $\begingroup$ @runway44 yes you are correct $\endgroup$ – sqrt-3299 Jul 11 at 21:17
  • $\begingroup$ Related question; math.stackexchange.com/questions/3268069/… $\endgroup$ – Servaes Jul 11 at 22:25
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Apply $$x=(x-y)+y$$ and expand the binomial and cancel and you get :$$(x-y)((x-y)^{p-1}+p(x-y)^{p-2}y+\cdots+py^{p-1})$$ The factorization is conditional. It has the gcd of $x-y$ and $y$ raised to $p-1$ as a factor if p is in their gcd, that means $p$ can be raised to at least the $p$ power.

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